# Find Partial Sum Formula of a serie

I have such question on my calculus exercises :

Find the sum for $$\displaystyle T_n = \sum_{k=-3}^{n+2} \left(\frac{-3}{2}\right)^{k+1}$$

To be honest, I don't really have an idea how to do this. However, this does resemble a little bit to Geometric Serie (But geometric series shouldn't really start with a negative index).

This is the procedure from the answer key which I don't understand how they transform and eventually use the Partial Sum Formula for Geometric Serie:

1. $$\displaystyle T_n = \sum_{k=-3}^{n+2} \left(\frac{-3}{2}\right)^{k+1}$$ (Geometric series with radius of $$\displaystyle \frac{-3}{2}$$)
2. $$\displaystyle T_n = \left(\frac{-3}{2}\right)^{-3+1} \cdot \frac{1-\left(\frac{-3}{2}\right)^{n+2-(-3)+1}}{1-\left(\frac{-3}{2}\right)}$$
3. $$\displaystyle T_n = \frac{9}{10}\left(1-\left(\frac{-3}{2}\right)^{n+6}\right)$$

Can someone help me explain what is exactly happening here? It is very confusing for me

Re-index ($$m=k+3$$) the geometric sum as $$T_n=\sum_{k=-3}^{n+2}\left(\frac{-3}{2}\right)^{k+1}=\sum_{m=0}^{n+5}\left(\frac{-3}{2}\right)^{m-2}=\sum_{m=0}^{n+5}\left(\frac{-3}{2}\right)^{-2}\left(\frac{-3}{2}\right)^{m}$$ and apply the standard formula for a geometric sum ($$q\neq1$$): $$S=\sum_{k=0}^{N}a_0q^{k}=a_0\frac{1-q^{N+1}}{1-q}$$ with $$a_0=\left(\frac{-3}{2}\right)^{-2}$$ and $$q=\frac{-3}{2}$$ and $$N=n+5$$.
This exercise is testing your ability to adjust the indices before applying the default geometric series formulae. If you think about what the sum itself means, you can separate the negative terms and then add on the additional terms after using your fomulae on the sum. However, the better way to do it is to adjust the indices by factoring out the appropriate power of -3/2, until the sum starts from the $$0^{th}$$ power as usual. $$T_n=\sum_{k=-3}^{n+2} \left( \dfrac{-3}{2}\right)^{k+1}\\ = \left( \dfrac{-3}{2}\right)^{-2}\Bigg( \sum_{k=-3}^{n+2} \left( \dfrac{-3}{2}\right)^{k+3} \Bigg)\\ = \left( \dfrac{-3}{2}\right)^{-2}\Bigg( \sum_{k=0}^{n+5} \left( \dfrac{-3}{2}\right)^{k} \Bigg)\\ = \left( \dfrac{-3}{2}\right)^{-2}\ \dfrac{\dfrac{-3}{2}^{n+7}-1} {\dfrac{-3}{2}-1}.$$